Proving a Hausdorff space is a $T_1$-space

Question

Show that any $T_2$-space is also a $T_1$-space.

Let's consider $(X,\tau)$ to be a $T_2$ or Hausdorff topological space. Therefore $\exists x,y\in X$ such that $x\in U$, $y\in V$ and $U\cap V=\emptyset$. If it is defined $Y=\{x,y\}$ there exists the topology $\tau_y=\{A\cap Y:A\in\tau\}$, such that $(Y,\tau_y)$ is topological subspace of $(X,\tau)$. As $x\in U$, then $U\cap Y=U\cap{x,y}={y}$ then $y\in\tau_y$ so that $x=Y\setminus \{y\}$, is then a closed set of the topological subspace $(Y,\tau_y)$. Then $\{x\}\cap \{Y\}={x}\cap\{x,y\}={x}$, so that $\{x\}$ is closed in $X,\tau$(since the intersection of a closed set in Y produces a closed set in the respective subspace). In an analogous way it is show that $\{y\}$ is closed by the use of $V$. Since $x,y\in X$ are arbitrary then $(X,\tau)$ is $T_1$ space

Question:

Is my proof right?

Thanks in advance!

Answer 1

I'd use this characterisation, you just showed. If $X$ is Hausdorff and $x \neq y$ we have two open sets $U$ and $V$, with $x \in U, y \in V, U \cap V = \emptyset$. Then $V \cap \{x,y\} = \{y\}$ is clear, so $X$ is $T_1$.

Answer 2

I propose what follows. Suppose that $\{x\}$ is not closed. Then there exists $y\ne x$ in the closure of $\{x\}$. Next, by $T_2$, there is an open neighbourhood $U$ of $y$ s.t $x\not\in U$. It contradicts our conjecture $y\in\overline{\{x\}}.$